Question

A 25.0 g sample of water is placed in a calorimeter and left to stand. After 5 minutes, the calorimeter and water equilibrated at a temperature of 19.5 oC. A sample of boiling water was added to the calorimeter and the maximum temperature reached after mixing was 58.2 oC. What mass of boiling water was added to the cool water in the calorimeter? (The specific heat capacity of water is 4.184 J/g oC and the heat capacity of the calorimeter is 30.0 J/ oC).

Answer 1

The mass of boiling water added to the cool water in the calorimeter is approximately 4.21 g.

Define the variables:

m_water: mass of the cool water (25.0 g)

T_initial: initial temperature of the cool water and calorimeter (19.5 °C)

T_final: final temperature after mixing (58.2 °C)

c_water: specific heat capacity of water (4.184 J/g °C)

c_calorimeter: heat capacity of the calorimeter (30.0 J/°C)

m_boiling_water: mass of the boiling water (unknown)

Calculate the heat gained by the cool water:

Q_water = m_water * c_water * (T_final - T_initial)

Substitute the values:

Q_water = 25.0 g * 4.184 J/g °C * (58.2 °C - 19.5 °C) = 1250.3 J

Calculate the heat lost by the calorimeter and the cool water:

Q_lost = Q_water + c_calorimeter * (T_final - T_initial)

Substitute the values:

Q_lost = 1250.3 J + 30.0 J/°C * (58.2 °C - 19.5 °C) = 1415.3 J

Since the heat is lost by the calorimeter and cool water, it is gained by the boiling water:

Q_gained = -Q_lost

Calculate the mass of the boiling water:

m_boiling_water = Q_gained / (c_water * 100)

Since the boiling water is at 100 °C, the initial temperature difference is 80.5 °C.

Substitute the values:

m_boiling_water = -1415.3 J / (4.184 J/g °C * 80.5 °C) = -4.21 g = 4.21 g